Problem 34 - Precalculus Mathematics for Calculus

Question

Find the complete solution of the linear system, or show that it is inconsistent. $$\left\\{\begin{aligned} 2 x+4 y-z &=3 \\ x+2 y+4 z &=6 \\ x+2 y-2 z &=0 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The system has infinitely many solutions: \((x, y, z) = (2 - 2k, k, 1)\) for \( k \in \mathbb{R} \).

Step 1: Write the System in Matrix Form

The given system of equations can be expressed as an augmented matrix. For the system \( 2x + 4y - z = 3 \), \( x + 2y + 4z = 6 \), and \( x + 2y - 2z = 0 \), the corresponding augmented matrix is:\[\begin{bmatrix}2 & 4 & -1 & | & 3 \1 & 2 & 4 & | & 6 \1 & 2 & -2 & | & 0\end{bmatrix}\]

Step 2: Perform Row Operations to Achieve an Upper Triangular Form

Apply row operations to transform the matrix into an upper triangular form. First, let's eliminate the first entry in the second row ("1") by using the first row:Subtract \( \frac{1}{2} \times \text{Row 1} \) from \( \text{Row 2} \):\[\begin{bmatrix}2 & 4 & -1 & | & 3 \0 & 0 & 4.5 & | & 4.5 \1 & 2 & -2 & | & 0\end{bmatrix}\]Next, eliminate the first entry in the third row ("1") using the first row:Subtract \( \frac{1}{2} \times \text{Row 1} \) from \( \text{Row 3} \):\[\begin{bmatrix}2 & 4 & -1 & | & 3 \0 & 0 & 4.5 & | & 4.5 \0 & 0 & -1.5 & | & -1.5\end{bmatrix}\]

Step 3: Further Simplify the Triangular Matrix

To achieve a better understanding, simplify the third row by performing suitable row operations:Multiply \( \text{Row 2} \) and \( \text{Row 3} \) by \(-1 \) to have positive coefficients:\[\begin{bmatrix}2 & 4 & -1 & | & 3 \0 & 0 & 4.5 & | & 4.5 \0 & 0 & 1.5 & | & 1.5\end{bmatrix}\]

Step 4: Interpret the Upper Triangular Matrix

Observe the obtained upper triangular matrix:From \( \text{Row 2} \) and \( \text{Row 3} \), we have:\[ 4.5z = 4.5 \text{ and } 1.5z = 1.5 \]Thus, \( z = 1 \).Substitute \( z = 1 \) back into the original second and third rows in order to solve for \( x \) and \( y \).

Step 5: Back-Substitute to Solve for Other Variables

Start with the equation from Row 1:\( 2x + 4y - 1 = 3 \)This simplifies to:\( 2x + 4y = 4 \)This means \( x + 2y = 2 \).Substitute this into the original Row 2:\( x + 2y + 4z = 6 \)\( x + 2y + 4 = 6 \)\( x + 2y = 2 \)Thus, \( x + 2y = 2 \) confirms the relation and the system has a solution for any \( y \).Setting \( y = k \), let \( x = 2 - 2y \), which is \( x = 2 - 2k \).

Step 6: Present the General Solution

Since the system allows freedom for the variable \( y \), a general solution is presented given that \( z = 1 \). The solution can be expressed as:\[\left( x, y, z \right) = \left( 2 - 2k, k, 1 \right) \quad \text{where } k \in \mathbb{R}\]This indicates the solution is a line of solutions in 3D space as \( k \) varies.

Key Concepts

Matrix Form

Transforming a system of linear equations into matrix form is a crucial first step in solving them efficiently. For the system of equations given:

• \( 2x + 4y - z = 3 \)
• \( x + 2y + 4z = 6 \)
• \( x + 2y - 2z = 0 \)

We translate this into an augmented matrix. The idea is to represent the coefficients of variables and their constants in a structured table format. Each row of the matrix corresponds to an equation, while each column corresponds to the coefficients of variables \(x\), \(y\), \(z\), followed by a column for the constants, separated by a vertical line:\[\begin{bmatrix} 2 & 4 & -1 & \vert & 3 \ 1 & 2 & 4 & \vert & 6 \ 1 & 2 & -2 & \vert & 0 \end{bmatrix}\]This matrix form not only neatly organizes the information but also sets the stage for performing further operations systematically.

Row Operations

Row operations are mathematical manipulations used to simplify matrices and reveal solutions to systems of equations. They include:

• Swapping the positions of two rows (Row Interchange)
• Multiplying an entire row by a non-zero constant (Row Scaling)
• Adding or subtracting multiples of one row to another (Row Addition/Subtraction)

These operations are aimed at transforming the matrix into a simpler form, such as an upper triangular matrix, from which solutions can be more readily identified. In this exercise, we applied row operations to eliminate terms below the diagonal. For instance, we eliminated the first entry in the second row by subtracting half of the first row from the second row, thus modifying the entries to create zeros below the leading coefficient in the first row. Through these operations, we steadily progress toward our goal of simplifying the system.

Upper Triangular Matrix

An upper triangular matrix is a type of matrix where all entries below the main diagonal are zero. This form is beneficial for solving linear systems because it simplifies back substitution. In our previous steps, we modified the original matrix into this more manageable form:\[\begin{bmatrix} 2 & 4 & -1 & | & 3 \ 0 & 0 & 4.5 & | & 4.5 \ 0 & 0 & 1.5 & | & 1.5 \end{bmatrix}\]With an upper triangular matrix, the focus is on systematically reducing non-essential information or terms below the rows of interest. Here, the transformation makes it immediately apparent that the third equation simplifies to \( 1.5z = 1.5 \), helping us solve for \( z = 1 \). This elimination method simplifies the equations, revealing solutions in a straightforward manner.

General Solution

Once the matrix reaches a form from which variable values can be directly extracted, we use back substitution to find exact solutions or identify if there are infinite solutions. In this exercise, solving the upper triangular matrix led to finding that \( z = 1 \). By substituting \( z \) back into earlier forms of the equations, further variables can be resolved.In the final refinement step, we see the relationship \( x + 2y = 2 \), holding true for any \( y \). Thus, the system provides a general solution:\[(x, y, z) = (2 - 2k, k, 1) \quad \text{where } k \in \mathbb{R}\]This general solution signifies a line of solutions in three-dimensional space. The parameter \( k \) represents any real number and specifies the degree of freedom the system holds, indicating solutions that are not unique but exist on a linear path due to the variable \( y \). This illustrates the beauty of linear algebra in demonstrating how such freedoms manifest in solution spaces.